Answer

Verified

348.8k+ views

Hint: First, analyse the odd and even terms separately from an A.P with common difference, from that of the original A.P. Compute the sum of odd and even terms and find the ratio in terms of n.

Complete step-by-step answer:

The A.P. containing 2n + 1 terms can be divided into two different A.P.s by separating the odd and even terms of the original A.P. The A.P. formed by taking the odd terms will have (n + 1) terms and the A.P. formed by taking the even terms will have n terms.

Let the first term of the original A.P. be a and the common difference be d. Then, the first odd term is a and the first even term is a + d.

We know the formula for the \[{n^{th}}\] term of the A.P.

\[{t_n} = a + (n - 1)d........(1)\]

The last even term of the A.P. is the \[2{n^{th}}\] term and it is given by:

\[{l_{even}} = a + (2n - 1)d.........(2)\]

The last odd term of the A.P. is the \[{(2n + 1)^{th}}\] term and it is given by:

\[{l_{odd}} = a + (2n + 1 - 1)d\]

\[{l_{odd}} = a + 2nd...........(3)\]

We know that the sum to n terms of the A.P. is given as:

\[{S_n} = \dfrac{n}{2}(a + l)............(4)\]

The sum of odd terms of the A.P. is given as:

\[{S_{odd}} = {S_{n + 1}} = \dfrac{{n + 1}}{2}(a + {l_{odd}})\]

From equation (3), we get:

\[{S_{odd}} = \dfrac{{n + 1}}{2}(a + a + 2nd)\]

\[{S_{odd}} = \dfrac{{n + 1}}{2}(2a + 2nd)\]

Taking 2 out as common and cancelling with the denominator, we have:

\[{S_{odd}} = (n + 1)(a + nd).........(5)\]

The sum of even terms of the A.P. is given as:

\[{S_{even}} = {S_n} = \dfrac{n}{2}(a + d + {l_{even}})\]

Substituting equation (2) in the above equation, we have:

\[{S_{even}} = \dfrac{n}{2}(a + d + a + (2n - 1)d)\]

\[{S_{even}} = \dfrac{n}{2}(2a + 2nd)\]

Taking 2 out as common and cancelling with the denominator, we have:

\[{S_{even}} = n(a + nd).........(6)\]

Dividing equation (5) by equation (6), we have:

\[\dfrac{{{S_{odd}}}}{{{S_{even}}}} = \dfrac{{(n + 1)(a + nd)}}{{n(a + nd)}}\]

Cancelling (a + nd), we have:

\[\dfrac{{{S_{odd}}}}{{{S_{even}}}} = \dfrac{{n + 1}}{n}\]

Hence, we proved that the ratio of the sum of odd terms and the sum of even terms is (n + 1):n

Note: You must be careful when you choose the first even terms of the A.P, it is a + d and not a. The number of odd terms is (n + 1) and not n.

Complete step-by-step answer:

The A.P. containing 2n + 1 terms can be divided into two different A.P.s by separating the odd and even terms of the original A.P. The A.P. formed by taking the odd terms will have (n + 1) terms and the A.P. formed by taking the even terms will have n terms.

Let the first term of the original A.P. be a and the common difference be d. Then, the first odd term is a and the first even term is a + d.

We know the formula for the \[{n^{th}}\] term of the A.P.

\[{t_n} = a + (n - 1)d........(1)\]

The last even term of the A.P. is the \[2{n^{th}}\] term and it is given by:

\[{l_{even}} = a + (2n - 1)d.........(2)\]

The last odd term of the A.P. is the \[{(2n + 1)^{th}}\] term and it is given by:

\[{l_{odd}} = a + (2n + 1 - 1)d\]

\[{l_{odd}} = a + 2nd...........(3)\]

We know that the sum to n terms of the A.P. is given as:

\[{S_n} = \dfrac{n}{2}(a + l)............(4)\]

The sum of odd terms of the A.P. is given as:

\[{S_{odd}} = {S_{n + 1}} = \dfrac{{n + 1}}{2}(a + {l_{odd}})\]

From equation (3), we get:

\[{S_{odd}} = \dfrac{{n + 1}}{2}(a + a + 2nd)\]

\[{S_{odd}} = \dfrac{{n + 1}}{2}(2a + 2nd)\]

Taking 2 out as common and cancelling with the denominator, we have:

\[{S_{odd}} = (n + 1)(a + nd).........(5)\]

The sum of even terms of the A.P. is given as:

\[{S_{even}} = {S_n} = \dfrac{n}{2}(a + d + {l_{even}})\]

Substituting equation (2) in the above equation, we have:

\[{S_{even}} = \dfrac{n}{2}(a + d + a + (2n - 1)d)\]

\[{S_{even}} = \dfrac{n}{2}(2a + 2nd)\]

Taking 2 out as common and cancelling with the denominator, we have:

\[{S_{even}} = n(a + nd).........(6)\]

Dividing equation (5) by equation (6), we have:

\[\dfrac{{{S_{odd}}}}{{{S_{even}}}} = \dfrac{{(n + 1)(a + nd)}}{{n(a + nd)}}\]

Cancelling (a + nd), we have:

\[\dfrac{{{S_{odd}}}}{{{S_{even}}}} = \dfrac{{n + 1}}{n}\]

Hence, we proved that the ratio of the sum of odd terms and the sum of even terms is (n + 1):n

Note: You must be careful when you choose the first even terms of the A.P, it is a + d and not a. The number of odd terms is (n + 1) and not n.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

How many crores make 10 million class 7 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths